Simplify; express your answer in exponential form. Assume $q\neq 0, x\neq 0$. $\dfrac{{(qx^{-1})^{3}}}{{(q^{-4}x^{2})^{2}}}$
To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(qx^{-1})^{3} = (q)^{3}(x^{-1})^{3}}$ On the left, we have ${q}$ to the exponent ${3}$ . Now ${1 \times 3 = 3}$ , so ${(q)^{3} = q^{3}}$ Apply the ideas above to simplify the equation. $\dfrac{{(qx^{-1})^{3}}}{{(q^{-4}x^{2})^{2}}} = \dfrac{{q^{3}x^{-3}}}{{q^{-8}x^{4}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{3}x^{-3}}}{{q^{-8}x^{4}}} = \dfrac{{q^{3}}}{{q^{-8}}} \cdot \dfrac{{x^{-3}}}{{x^{4}}} = q^{{3} - {(-8)}} \cdot x^{{-3} - {4}} = q^{11}x^{-7}$